When $n=1$, the left hand size evaluates to $$\sum_{k=1}^1\sum_{j=1}^k \frac{a_j}{k}\binom{k}{j}$$ which simplifies to:

When $n=1$, the right hand side evaluates to $$\sum_{k=1}^1\frac{a_k}{k}\binom{1}{k}$$ which simplifies to:

Great, so we have proved the base case. Now assume the statement holds for $n$, i.e., we will assume $$\sum_{k=1}^n\sum_{j=1}^k \frac{a_j}{k}\binom{k}{j}=\sum_{k=1}^n \frac{a_k}{k}\binom{n}{k}$$ We will use this assumption to show the identity is true for $n+1$. Notice that $$\sum_{k=1}^{n+1}\sum_{j=1}^k\frac{a_j}{k}\binom{k}{j}=\sum_{k=1}^n\sum_{j=1}^k\frac{a_j}{k}\binom{k}{j}+\sum_{\ell=1}^{n+1}\frac{a_\ell}{X}\binom{X}{\ell}$$ what is $X$?

Notice that the first sum on the right hand side is equal to $$\sum_{k=1}^n \frac{a_k}{k}\binom{n}{k}$$ by induction. Therefore $$\sum_{k=1}^{n+1}\sum_{j=1}^k \frac{a_j}{k}\binom{k}{j}=\sum_{k=1}^n \frac{a_k}{k}\binom{n}{k}+\sum_{\ell=1}^{n+1}\frac{a_\ell}{n+1}\binom{n+1}{\ell}.$$ Note that $$\frac{a_\ell}{n+1}\binom{n+1}{\ell}=\frac{a_\ell}{\ell}\binom{X}{\ell-1}$$ what is $X$?

Using the identity given above, we can see $$\sum_{k=1}^n \frac{a_k}{k}\binom{n}{k}+\sum_{\ell=1}^{n+1}\frac{a_\ell}{n+1}\binom{n+1}{\ell}=\sum_{k=1}^n\frac{a_k}{k}\left(\binom{n}{k}+\binom{n}{k-1}\right)+\frac{a_X}{X}\binom{n+1}{n+1}$$ what is $X$?

Which named identity can we use to simplify $$\binom{n}{k}+\binom{n}{k-1}?$$

Therefore we have $$\sum_{k=1}^{n+1}\sum_{j=1}^k \frac{a_j}{k}\binom{k}{j}=\sum_{k=1}^n \frac{a_k}{k}\binom{n+1}{k}+\frac{a_{n+1}}{n+1}\binom{n+1}{n+1}=\sum_{k=1}^{n+1} \frac{a_k}{k}\binom{n+1}{k}$$ So by what can we conclude the identity holds? (Hint: The answer is induction.)